Lecture13

Lecture13 - Lecture 13 In (15) of Lecture 11, the...

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Unformatted text preview: Lecture 13 In (15) of Lecture 11, the electrical equations of an electric machine are put in the first- order standard form: ) , ( u x f dt x d = (1) where u is a vector of the voltages and the load torque T L , = m i x and where = R R S S i i i i i . Specifically in the - frame, the equations are: } ] [ ] [ ] [ { ] [ 1 i F i G n i R e L i p m = (2) 1 1 / )} ( { J i i i i nM T f p R S R S L m m + = (3) In general, the equations are solved by numerical integration. It is important to understand the concept the state ) ( t x at time t in state-space. Specifying the state ) ( t x at time t means that it is not necessary to know the history of ) ( t x for t t . On the other hand, the initial condition ) ( t x at time t determines the solution of ) , ( u x f dt x d = for t t . The integration software requires the initial state because without it, it cannot evaluate ) ), ( ( u t x f dt x d = , the initial gradient to start the integration. Steady-State Solutions - frame In the - frame, because u is time invariant, one expects that X t x = ) ( is also time invariant. As such, the steady-solution satisfies = dt x d , which means that ) , ( = u x f . This can be reached numerically integrating (2) and (3) long enough until a time invariant solution X t x = ) ( is reached....
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Lecture13 - Lecture 13 In (15) of Lecture 11, the...

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